Question

find y" of xy +tan exponent of-1(xy) =4

Answer 1

is that find 2nd derivative?

Answer 2

yes could you help me?

Answer 3

well you need to find 1st derivative first using implicit differentiation
\[d/dx \tan^{-1} u = \frac{1}{1+u^{2}}*du\]
where u = xy in this case
(xy)' = y + xy'
\[\rightarrow y+xy' +\frac{y+xy'}{1+x^{2}y^{2}} = 0\]
\[(y+xy')(1+x^{2}y^{2}) + (y+xy') = 0\]
\[\rightarrow y+xy' = 0\]
\[y' =- \frac{y}{x}\]
Now differentiate again using quotient rule
\[y'' = \frac{-xy'+y}{x^{2}}\]
substitute in for y'
\[y'' = \frac{2y}{x^{2}}\]